- #1
Yitzach
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I have in mind to build a game to help teach or demonstrate some concepts in QM and I thought it would be nice to be able to measure momentum. So as a proof of concept before I get too many man hours burned on the project I thought it would be good to do the infinite square well. I managed to get the un-normalized momentum probability out of my math. The only problem with it was that the integral on the real number line did not converge due to it being sinusoidal with a constant additive that gave periodic zeros in the trough. So then I think to integrate between -c and +c. But as this is a classical system, c is not a necessary universal speed limit. So now I would like the relativistic momentum operator. I'm not seeing that available, yet. So using Griffths's book on QM as my resource:
[itex]\hat{p}=\frac{mv}{\sqrt{1-(\frac{v}{c})^2}}[/itex]6.47
[itex]p=-i\hbar\nabla\Rightarrow v=-\frac{i\hbar}{m}\nabla[/itex]cover
[itex]\hat{p}=\frac{-i\hbar\nabla}{\sqrt{1-(\frac{-i\hbar\nabla}{mc})^2}}[/itex]
[itex]\hat{p}=\frac{-i\hbar\nabla}{\sqrt{1+(\frac{\hbar}{mc})^2\nabla^2}}[/itex]
Use:
[itex]\int\Psi*\frac{-i\hbar\nabla}{\sqrt{1+(\frac{\hbar}{mc})^2\nabla^2}}\Psi dx=\left<p\right>[/itex]
[itex]\int\Psi*\frac{-i\hbar\nabla\Psi}{\sqrt{1+(\frac{\hbar}{mc})^2\nabla^2\Psi}} dx=\left<p\right>[/itex]
This doesn't look quite intelligent to me. Does anyone have a better solution to the problem? It would be wonderful if you could provide the eigenfuction for it as well. If this is it, the next trick will be to solve the following equation or an approximation of it:
[itex]\frac{-i\hbar\nabla\Psi}{\sqrt{1+(\frac{\hbar}{mc})^2\nabla^2\Psi}}=p\Psi[/itex]
[itex]\hat{p}=\frac{mv}{\sqrt{1-(\frac{v}{c})^2}}[/itex]6.47
[itex]p=-i\hbar\nabla\Rightarrow v=-\frac{i\hbar}{m}\nabla[/itex]cover
[itex]\hat{p}=\frac{-i\hbar\nabla}{\sqrt{1-(\frac{-i\hbar\nabla}{mc})^2}}[/itex]
[itex]\hat{p}=\frac{-i\hbar\nabla}{\sqrt{1+(\frac{\hbar}{mc})^2\nabla^2}}[/itex]
Use:
[itex]\int\Psi*\frac{-i\hbar\nabla}{\sqrt{1+(\frac{\hbar}{mc})^2\nabla^2}}\Psi dx=\left<p\right>[/itex]
[itex]\int\Psi*\frac{-i\hbar\nabla\Psi}{\sqrt{1+(\frac{\hbar}{mc})^2\nabla^2\Psi}} dx=\left<p\right>[/itex]
This doesn't look quite intelligent to me. Does anyone have a better solution to the problem? It would be wonderful if you could provide the eigenfuction for it as well. If this is it, the next trick will be to solve the following equation or an approximation of it:
[itex]\frac{-i\hbar\nabla\Psi}{\sqrt{1+(\frac{\hbar}{mc})^2\nabla^2\Psi}}=p\Psi[/itex]